A is the set of ten consecutive numbers, while the average of second lowest and the highest number of set A is 32. If B is the set of five consecutive odd number which lowest number is equal to second highest number of set A, then find the average of set B?

The correct answer is 39.

We solve this in two stages: first find Set A, then use it to define and evaluate Set B.

Stage 1: Finding Set A

Let the lowest (smallest) number in Set A be n. Since A contains 10 consecutive numbers, the full set is:

n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9

From this:

- Second lowest number = n + 1
- Highest number = n + 9

We are told the average of the second lowest and the highest is 32:

\frac{(n+1)+(n+9)}{2}=32

\frac{2n+10}{2}=32

n+5=32

n=27

So Set A is: 27, 28, 29, 30, 31, 32, 33, 34, 35, 36

The second highest number of Set A (i.e., second from the top) = 35

Stage 2: Finding Set B and its Average

Set B contains 5 consecutive odd numbers, and its lowest number = 35 (the second highest of Set A).

So Set B is: 35, 37, 39, 41, 43

The average of Set B:

\text{Average}=\frac{35+37+39+41+43}{5}=\frac{195}{5}=39

Notice that for any set of consecutive odd numbers, the average is simply the middle term. Here, the middle (3rd) term of Set B is 39, which confirms our answer.

Therefore, the average of Set B = 39.